) If 𝑓 (𝑥, 𝑦) = 𝑒
𝑥𝑦, where 𝑥 = 2𝑡4+1
and 𝑦 = 3𝑡 + 5. Find the total derivative dd𝑡𝑓
at 𝑡 = 0.

To find the total derivative \( \frac{d}{dt} f \) at \( t = 0 \), where \( f(x, y) = e^{xy} \), \( x = 2t^4 + 1 \), and \( y = 3t + 5 \), we can apply the chain rule for multivariable functions. 

The total derivative of \( f \) with respect to \( t \) is given by:
\[
\frac{d}{dt} f(x, y) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}
\]

### Step 1: Calculate \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \)
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}
\]
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( e^{xy} \right) = x e^{xy}
\]

### Step 2: Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
Given:
\[
x = 2t^4 + 1 \quad \text{and} \quad y = 3t + 5
\]
\[
\frac{dx}{dt} = \frac{d}{dt} \left( 2t^4 + 1 \right) = 8t^3
\]
\[
\frac{dy}{dt} = \frac{d}{dt} \left( 3t + 5 \right) = 3
\]

### Step 3: Plug in the values of \( x \), \( y \), \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \)
At \( t = 0 \):
\[
x = 2(0)^4 + 1 = 1
\]
\[
y = 3(0) + 5 = 5
\]
\[
\frac{\partial f}{\partial x} = 5e^{1 \cdot 5} = 5e^5
\]
\[
\frac{\partial f}{\partial y} = 1e^{1 \cdot 5} = e^5
\]
\[
\frac{dx}{dt} = 8(0)^3 = 0
\]
\[
\frac{dy}{dt} = 3
\]

### Step 4: Compute the total derivative
\[
\frac{d}{dt} f = \left( 5e^5 \right) \cdot 0 + \left( e^5 \right) \cdot 3 = 3e^5
\]

Thus, the total derivative \( \frac{d}{dt} f \) at \( t = 0 \) is \( 3e^5 \).

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How is the chain rule applied to multivariable functions?
2. What are partial derivatives, and how are they different from total derivatives?
3. How does the value of the total derivative change if \( t \) is not zero?
4. Can you explain the relationship between partial derivatives and directional derivatives?
5. How would you interpret the result of this derivative in a real-world context?

**Tip:** Remember, the chain rule is a powerful tool for differentiating composite functions, especially when dealing with functions of multiple variables.

Learn how to find the total derivative of the function f(x, y) = e^(xy) at t = 0. This problem involves applying the chain rule for multivariable functions, calculating partial derivatives, and evaluating at a specific point. Suitable for college-level students.

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